1. Field of the Invention
The present invention relates to a switched capacitor circuit.
2. Description of Related Art
In a circuit that handles a high frequency signal, such as a cellular phone, it is necessary to match a characteristic impedance to transmit a signal power efficiently. Matching the characteristic impedance requires an accurate inductor and capacitor. However, a variation in capacitance value becomes a problem in a semiconductor circuit. Therefore, it is necessary to detect the variation in capacitance value. To detect the variation, both a reference capacitor and an object capacitor are configured to act as switched capacitors. Then, an operational amplifier is driven through these equivalent resistors, thereby detecting the variation in capacitance as a voltage.
FIGS. 3A and 3B show an integrator 1 using a switched capacitor circuit. As shown in FIGS. 3A and 3B, the integrator 1 includes an operational amplifier OP1, a switching capacitor C1, a feedback capacitor C2, and a switch SW1.
An operation of the integrator 1 is explained with reference to FIGS. 3A and 3B. As shown in FIG. 3A, the switch SW1 is connected to a side of an input terminal Vin. Assume that an input voltage Vin is applied to the input terminal Vin. At this time, the switching capacitor C1 is charged according to the voltage Vin.
Referring next to FIG. 3B, the switch SW1 is connected to a side of an inverting input terminal of the operational amplifier OP1. Because voltages at both terminals of the switching capacitor C1 are set to a ground voltage GND, the electric charge stored in the switching capacitor C1 is discharged. In this manner, the switching capacitor C1 is repeatedly charged and discharged by alternately repeating the connection states of FIGS. 3A and 3B. Assuming that a charge and discharge current is represented by Ij and a switching frequency of the switch SW1 is represented by Fs at this time, the following expression (1) holds.Ij=C1×Vin×Fs  (1)
As shown in the expression (1), a constant current flows to the switching capacitor C1. Thus, the switching capacitor C1 can be considered as an equivalent resistance. Therefore, the integrator 1 can be considered as an equivalent circuit like a circuit shown in FIG. 3C. A time constant T of this integrator 1 is represented by the following expression (2). The time constant T depends only on a relative accuracy regardless of an absolute variation of the capacitors C1 and C2.
                                                        T              =                              R                ⁢                                                                  ⁢                1                ×                C                ⁢                                                                  ⁢                2                                                                                        =                                                Vin                  Ij                                ×                C                ⁢                                                                  ⁢                2                                                                                        =                                                Vin                                      C                    ⁢                                                                                  ⁢                    1                    ×                    Vin                    ×                    Fs                                                  ×                C                ⁢                                                                  ⁢                2                                                                                        =                                                C                  ⁢                                                                          ⁢                  2                                                  C                  ⁢                                                                          ⁢                  1                  ×                  Fs                                                                                        (        2        )            
FIGS. 4A to 4C show a voltage amplification circuit 2 using a switched capacitor, in which the configuration of the integrator 1 shown in FIGS. 3A to 3B is changed. As shown in FIGS. 4A and 4C, the voltage amplification circuit 2 includes the operational amplifier OP1, the switching capacitor C1, the feedback capacitor C2, and switches SW1 to SW3.
An operation of the voltage amplification circuit 2 is explained with reference to FIGS. 4A and 4B. As shown in FIG. 4A, the switches SW1 and SW2 are connected to a side of the inverting input terminal of the operational amplifier ON, and the switch SW3 is connected to a side of an output terminal Vout. Assume that an output voltage of the output terminal Vout is represented by Vout. At this time, the capacitors C1 and C2 are charged.
Referring next to FIG. 4B, the switch SW1 is connected to the side of the inverting input terminal, and the switches SW2 and SW3 are connected to the side of the ground terminal GND. Therefore, the electric charge stored in the capacitors C1 and C2 are discharged. A charging current I1 flows to the capacitor C1 and a charging current I2 flows to the capacitor C2, due to the repetition of the charge and discharge. These currents I1 and I2 are expressed by the following expressions (3) and (4), respectively.I1=C1×Vin×Fs  (3)I2=C1×(Vout−Vin)×Fs  (4)
The capacitors C1 and C2 act as equivalent resistors as seen from the expressions (3) and (4). Then, the voltage amplification circuit 2 can be considered as an equivalent circuit like a circuit shown in FIG. 4C.
When the output voltage Vout at this time is represented by Vin, Expression (5) holds. The output voltage Vout is represented by a capacitance ratio between the capacitors C1 and C2.
                                                        Vout              =                              Vin                ×                                                                            R                      ⁢                                                                                          ⁢                      1                                        +                                          R                      ⁢                                                                                          ⁢                      2                                                                            R                    ⁢                                                                                  ⁢                    1                                                                                                                          =                              Vin                ×                                                                            Vin                                              I                        ⁢                                                                                                  ⁢                        1                                                              +                                                                  Vout                        -                        Vin                                                                    I                        ⁢                                                                                                  ⁢                        2                                                                                                  Vin                                          I                      ⁢                                                                                          ⁢                      1                                                                                                                                              =                              Vin                ×                                                                            1                                              C                        ⁢                                                                                                  ⁢                        1                        ×                        Fs                                                              +                                          1                                              C                        ⁢                                                                                                  ⁢                        2                        ×                        Fs                                                                                                  1                                          C                      ⁢                                                                                          ⁢                      1                      ×                      Fs                                                                                                                                              =                              Vin                ×                                  (                                      1                    +                                                                  C                        ⁢                                                                                                  ⁢                        1                        ×                        Fs                                                                    C                        ⁢                                                                                                  ⁢                        2                        ×                        Fs                                                                              )                                                                                                        =                              Vin                ×                                  (                                      1                    +                                                                  C                        ⁢                                                                                                  ⁢                        1                                                                    C                        ⁢                                                                                                  ⁢                        2                                                                              )                                                                                        (        5        )            
In the manner, assuming that the capacitor C2 is used as a reference capacitor, for example, a capacitor whose variation is to be detected is connected to the capacitor C1. This makes it possible to detect a variation in capacitance as a voltage.
Note that an integrator that uses a switched capacitor is disclosed in Japanese Unexamined Patent Application Publication No. 2003-203195. A technique in which a low-pass filter is connected in cascade with a switched capacitor filter is disclosed in Japanese Unexamined Patent Application Publication No. 58-198918.